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Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
A Guide to Solving NMR Problems
NMR spectroscopy is a great tool for determining structures of organic compounds.
As you know 1H spectra have three features, chemical shift, signal intensity, and
multiplicity, each providing helpful information. In this document we show how you use
these features together to assign structures from 1H and 13C spectra. Use this approach.
I Begin with a general examination of the spectrum:
a) determine how many signals (clusters) there are in the 1H spectrum and note their
chemical shifts, and count the signals in the 13C spectrum;
b) determine how many 1H nuclei are present in each cluster (look at the integration);
c) determine the multiplicity of each signal cluster (count the peaks or read the label
attached to the peak);
d) determine what additional information is available (often the molecular formula is
given; that gives you a chance to subtract groups you’ve identified and to
check what is left to identify);
e) compare the number of 1H signals with those in the 13C spectrum; this may tell
you whether all carbon atoms bear 1H nuclei or whether there are carbons
without 1H nuclei attached.
II Now proceed to the interpretation of the spectrum:
f) for each peak match the number of 1H nuclei and its multiplicity with a
corresponding peak and combine them to groups (coupling is mutual);
g) record each group you recognize (that’ll give you partial credit J );
h) when you have assigned all 1H groups that you recognize, add their formulas up
and compare with the molecular formula; the difference will be a group or
groups that you haven’t identified as yet or cannot see in the 1H spectrum;
step e) above will give you similar information.
i) enter the “difference” and combine all fragments.
III Check your work: make sure that you have accounted for all signals and for every
atom of the molecular structure and that your combined structure satisfies all
chemical shifts. That’s all, folks.
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
Lets begin by looking at a really quite simple compound; it has a molecular formula
of C4H6Cl2. The 1H spectrum has 3 signals and the 13C spectrum shows 4 signals; make a
note that there must be a C without any H attached.
tri
3H
s
1H
q
2H
10
8
6
PPM
4
2
0
For your interpretation of the individual peaks, it’s best to start at one of the edges,
low field or high, left or right. At low field (left) our spectrum has a singlet (1 H) at 9.7 ppm;
that chemical shift is unmistakable: it can only indicate an aldehyde. Because this signal is a
singlet (n + 1 = 1; n = 0), there cannot be any1H nuclei on the adjacent carbon. You can enter
the aldehyde fragment without neighbor in the “box provided” for your first piece of partial
credit. (I place groups with low-field signals on the left, but that is not crucial).
O
H
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
Now we go to high field (right): there is a triplet (3 H) at 0.9 ppm; 3 H at high field is
almost always a methyl group. The signal is a triplet (n + 1 = 3); therefore, the methyl group
must have (n =) 2 1H neighbors; that must be a CH2 group. The CH2 signal is a quartet (n +
1 = 4), so it must have (n =) 3 1H neighbors; that can only be the CH3 group. You also know
that there cannot be any 1H neighbor on the other side. Note that the combination of a
“triplet, 3 H” with a “quartet, 2 H” is ALWAYS a C2H5 group. You can enter an ethyl group
into the “box provided”, joining the aldehyde group (and increasing your partial credit).
O
CH2
H
CH3
IMPORTANT: it does NOT matter in what order we probe the spectrum. If we had
started with the high field region (C2H5 before CHO), the conclusion would be the same.
So far we have no information how the two groups are connected. The answer could
come from the composition or from a 13C spectrum. Working with the empirical formula,
C4H6Cl2, you subtract the two fragments, CHO and C2H5 (C4H6Cl2 – CHO – C2H5 = CCl2)
and are left with a dichloromethylene unit (more partial credit).
We find support for our structure fragments in the 13C spectrum. The four signals,
198 ppm for the aldehyde carbon, 102 ppm for the CCl2 carbon next to the C=O group, and
30 and 5 ppm, respectively, for the (slightly deshielded) CH2 and the (shielded) CH3 of the
ethyl group, are in accord with the fragments we identified. Note, that we base the CCl2
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
group on the 13C spectrum and use non-spectroscopic information (the composition) in order
to derive this part of the target structure.
There is only one way how the three units can be connected into a molecule.
O
H
O
H2
C
CH2
Cl
CCl2
CHO
H
CH3
Cl
CH3
Cl Cl
C2H5
At this point (or earlier, if you wish) you can use the composition to check the
degrees of unsaturation (Hsat = 2 x 4 + 2 = 10; – 2 for Cl2 = 8; Hact = 6; Hsat – Hact = 2;
D(U) = 1). The D(U) count is in accord with the presence of the C=O group. This gives you
additional confidence that your assignment is correct. So much for a simple spectrum.
Are you ready for a slightly more difficult spectrum with two more carbon atoms
and two additional signals each in 1H and 13C spectra? Its composition is C6H10Cl2O).
tri
2H
d
2H
tri
3H
tri
1H
sex
2H
10
8
6
PPM
4
2
0
This spectrum has a triplet (1 H) at 9.8 ppm, clearly indicating the presence of an
aldehyde group; because the signal is a triplet (n + 1 = 3), there must be two 1H nuclei on the
adjacent carbon (n = 2). We find a 2 H signal, a doublet, at 2.9 ppm, indicating that it has 1
1H neighbor (the coupling between groups is always mutual).
The group is slightly
deshielded, indicating that it resides near a group, like C=C or C=O. This suggests that we
combine the aldehyde function with the CH2 group and enter them into the “box provided”.
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
O
H
220
200
C
H2
180
160
140
120
100
PPM
80
60
40
20
0
Turning to the high-field side of the spectrum, there is a triplet (3 H) at 0.9 ppm; any
3 H at high field must be a methyl group; the triplet (n + 1 = 3) indicates that the methyl
group must have (n =) 2 1H neighbors, i.e., a CH2 group. There are two CH2 groups in the
spectrum, at 1.4 and 1.8 ppm; which one should we choose? The signal at 1.8 ppm is a triplet
(n + 1 = 3); attachment to the CH3 group (n = 3) would require at least a quartet. The CH2
signal is a sextet (n + 1 = 6), requiring 5 1H neighbors; since we have 3 neighbors at high
field (the CH3 group), we need 2 more at lower field, the triplet at 1.8 ppm. We combine
these three groups to a slightly deshielded propyl group, which we enter into the “box
provided”. Note: the combination of triplet (3H) – sextet (2H) – triplet (2H) is always a
propyl group. Once again the result is independent of the sequence: examining the high field
signals before the low field ones yields the same result as the sequence used above.
O
CH2
H
C
H2
C
H2
CH3
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
In order to determine how these groups are connected, we consult the composition
(C6H10Cl2O). If we subtract from the empirical formula, C6H10Cl2O, the fragments,
CHOCH2 and C3H8, we are left with CCl2, a dichloromethylene unit. Again, there is only
one way how the resulting three units can be connected into a molecule.
Cl Cl
O
H
C
H2
O
C
H2
CH2
CH3
H
Cl Cl
C
H2
C
H2
CH2
CH3
We check the 13C spectrum whether it fits the assigned structure. It has five signals: a
shift of 202 ppm is characteristic for a C=O group; 70 ppm for the CCl2 carbon (not directly
next to the C=O group as in our first spectrum); 58 and and 48 ppm for the deshielded CH2
groups next to C=O and CCl2; and 18 and 13 ppm for the shielded CH2 and CH3 belonging
to the propyl group. This 13C spectrum fully supports the structure we derived.
Checking the degrees of unsaturation (you could have done this earlier, if you
wanted): Hsat = 2 x 6 + 2 = 14; – 2 for Cl2 = 12; Hact = 10; Hsat – Hact = 2; D(U) = 1). The
presence of the C=O group is in accord with one element of unsaturation (the C=O) group.
This agreement gives you additional confidence that your assignment is correct.
So far we have dealt with spectra in which the signals of the structure fragments were
well separated. Maybe we should try to solve a problem where the signals of the structure
fragments fall into the same region. Let us consider the spectrum of a compound, whose
chemical formula is C6H12O2; the spectra, shown below, have five 1H and six 13C signals.
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
180
160
140
120
100
PPM
80
60
40
20
0
This time let’s start with the degrees of unsaturation (there is nothing magical about
the sequence): Hsat = 2 x 6 + 2 = 14; Hact = 12; Hsat – Hact = 2; D(U) = 1): that means we
have to expect one double bond or one ring.
Our 1H spectrum has two methyl groups, at 0.9 and 1.2 ppm; both are triplets (n + 1 =
3), so both must be adjacent to a CH2 group. We have three CH2 groups to choose from, at
1.7, 2.3, and 3.7 ppm; they are slightly to significantly deshielded. The group at 2.3 ppm is a
quartet (n + 1 = 4); obviously, it is located next to a methyl group. Combining the two gives
you an ethyl group – “bank” it.
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
CH2
CH3
Of the remaining three groups, the sextet (n + 1 = 6) at 1.7 ppm requires n = 5 1H
neighbors; it just so happens that we have one CH3 and one CH2 group (3.7 ppm) left. This
could amount to a propyl group, whose open-ended CH2 is significantly deshielded. As you
place it in the “box provided”, you may remember that the combination, triplet (3H) – sextet
(2H) – triplet (2H), is always a propyl group.
CH2
H3C
CH2
C
H2
CH3
How could these groups be connected? In order to answer that question we remember
the composition, C6H12O2, and subtract from it the two fragments C2H5 and C3H7, giving us
CO2; this represents an “ester” linkage, –O–C(=O)–. The C=O function takes care of the
degree of unsaturation we expected. [You will learn much more about esters in Chem 308.]
O
CH2
H3C
C
H2
CH2
O
CH3
In this case there are two ways to connect the three fragments: either the propyl or the
ethyl group could be attached to the O and the remaining group to the C=O. The answer lies
in the chemical shifts of the triplet at 3.7 ppm (part of the propyl group) and of the quartet at
2.3 ppm (part of the ethyl group). The shift of 3.7 ppm is exactly what we expect from a
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
group attached to the electronegative O; therefore we attach the propyl group to the O. 2.3
ppm is what we expect for a CH2 near a ( C=O ) double bond; accordingly, we attach the
ethyl group to the C=O group.
O
H2
C
H3C
CH2
CH3
O
CH2
Checking the 13C spectrum: an ester carbon has a very characteristic 13C chemical
shift (~175 ppm), clearly seen in the spectrum below. The other notable peak is the signal at
65 ppm, representing the CH2 carbon (of the propyl group), which is attached to the O. The
remaining five signals are unexceptional CH2 and CH3 resonances
One question remains (which does not affect the structure, but is still of interest):
which of the methyl signals belongs to the propyl and which belongs to the ethyl group. We
base this decision on chemical shift evidence: the CH3 group of the ethyl group is closer to a
deshielding unit (C=O); therefore we assign the signal at 1.2 ppm to that group and the triplet
at 0.9 ppm to the CH3 group of the propyl group. There, we have assigned a structure and
identified every signal unambiguously.
We hope that working through these three sets of spectra with additional supporting
information gives you some familiarity with the methodology and confidence to approach the
spectra you will encounter in your future exams. Good luck.
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth
Typical 1H and 13C Chemical Shift in Organic Molecules
Chemical Shift
Type of Hydrogen
0.8 – 1.0
Primary alkyl, RCH3
1.2 – 1.4
Secondary alkyl, RCH2R’
1.4 – 1.7
Tertiary alkyl, R3CH
1.6 – 1.9
Allylic, R2C=CR’CH3
2.2 – 2.5
Benzylic, ArCH2R
2.1 – 2.6
Next to carbonyl, R(C=O)CH2R’
1.7 – 3.1
≡ CH
Alkyne, RC
3.4 – 3.6
RCH2Br
3.6 – 3.8
RCH2Cl
3.3 – 4.0
RCH2OH, RCH2OR’
0.5 – 5.0
OH, SH, NH2
4.6 – 5.0
Terminal alkene, RHC=CH2
Terminal alkene near electronegative atoms
– 5.6
5.2 – 5.7
– 6.5
6.0 – 9.5
9.5 – 9.9
Internal alkene RHC=CH2
Internal alkene near electronegative atoms
Benzene and aromatics, ArH
Aldehyde, R(C=O)H
Chemical Shift
Type of Carbon
5 – 20
Primary alkyl, RCH3
20 – 30
Secondary alkyl, RCH2R’
30 – 50
Tertiary alkyl, R3CH
30 – 45
Quaternary alkyl, R4C
20 – 40
Allylic, R’2C=CRCH3
20 – 40
RCH2Br
25 – 50
RCH2Cl
50 – 90
RCH2OH, RCH2OR’
65 – 95
Alkyne, RC CH
Alkene, Aromatic
Ester, R(C=O)OR’
Aldehyde or Ketone, R(C=O)H , R(C=O)R’
100 – 160
170 - 175
190 – 210
≡
Organic Chemistry 307 – Solving NMR Problems – H. D. Roth